1. Title : Usage : Cone and Pyramid Subject : Mathematics
Target Audience :
Form 3
Usage :
Lecturing

2. Prerequisite knowledge :
1. Pythagoras’ Theorem.
2. Area of some plane figures
e.g. square, rectangle, triangle, circle, sector
3. Ratio and proportion
Objectives :
Let the students know and apply the mensuration concepts

3. Pyramid 1
Egyptian Pyramid

4. These are pyramids

5. V 
Vertex  
Slant edges
height   A B C D

6. Volume of pyramid A x
cube
Three congruent pyramids

7. x For any pyramid, Volume of pyramid =  base area  height
Volume of the pyramid = x3 x
x2 x = =
 base area  height
For any pyramid,
Volume of pyramid =
 base area  height

8. Example 1 cm Volume of the pyramid
The figure shows a pyramid with a rectangular base ABCD of area 192 cm2, VE = 15 cm and EF = 9 cm, find the volume of the pyramid. V A C D B
15 cm
9 cm E F V E F
15 cm
9 cm
Solution :
VF2 = ( ) cm2
VF = cm
= 12 cm
Volume of the pyramid =
 base area  height
= (
 192  12) cm3
= 768 cm3

9. Volume of a Frustum of a PyramidB
Pyramid A
 Pyramid B
 A frustum - =
= Volume of Pyramid A -
Volume of the frustum
Volume of Pyramid B

10. Example 2 Solution : = ( = ( V A E D C B H G F  ( 8  8 )  6) cm3
The base ABCD and upper face EFGH of the frustum are squares of side 16 cm and 8 cm respectively. Find the volume of the frustum ABCDEFGH. V A E D C B H G F
6 cm
12 cm
Solution :
= (
 ( 8  8 )  6) cm3
Volume of VEFGH
= cm3
= (
 ( 16  16 )  12) cm3
Volume of VABCD
= cm3
Volume of frustum ABCDEFGH
= ( ) cm3
= 896 cm3

11. Total surface area of a pyramidV B A D C V D C B A

12. + lateral faces Base Base area
Total surface area of pyramid VABCD = +
lateral faces
Base
Base area
The sum of of the area of all lateral faces + =
Total surface area of a pyramid

13. Example 3 Solution : V D C B A Total surface area of pyramid VABCD
The figure shows a pyramid with a rectangular base ABCD of area 48 cm2. Given that area of  VAB = 40 cm2 , area of  VBC = 30 cm2, find the total surface area of the pyramid.
Solution :
Total surface area of pyramid VABCD
= Area of ABCD
+ (Area VAB + Area VDC +
Area VBC + Area VAD )
= Area of ABCD
+ (Area VAB  2) +
(Area VBC  2)
= 48 cm2
+ ( (40  2) + (30  2)) cm2
= 188 cm2

14. How to generate a cone?
…...
…...

15. How to calculate the curved surface area ?
Cut here

16. Curved surface area = πr lθ r l
After cutting the cone,
Curved surface area = Area of the sector
Curved surface area = 1/2 ( l ) ( 2π r )
= π r l
Curved surface area = πr l

17. Volume of a cone r r h h
Volume of a cone = πr2 h 1 3

18. How to calculate total surface area of a cone?+ r
Total surface area =πr2 + πr l

19. Examples 1 a) If h = 12cm, r= 5 cm, what is the volume? Answer:
Volume = πr2h 1 3 1 3
= π (52) ( 12)
= 314 cm3

20. b) what is the total surface area?
= π52
Based Area
= 25πcm2
Slant height
= 
= 13 cm
Curved surface area
= π(5) ( 13)
= 65π cm2
Total surface area
= based area + curved surface area
= 25π+65π= 90π
= 282.6cm2 (corr.to 1 dec.place)

21. Volume of Frustum

22. Volume of Frustum = -  R r
Volume of frustum = volume of big cone - volume of small cone
= πR π r3
π( R3 - r3 ) =

23. Exercises
Start Now
Exit

24. Q1
8cm
The volume of a pyramid of square base is 96 cm3. If its height is 8 cm, what is the length of a side of the base?
A. 2 cm
B cm
C. 6cm
D. 12cm
E. 36cm
Answer
Answer is C
Help
To Q2

25. Q2 A X Y B C
In the figure, the volumes of the cone AXY and ABC are 16 cm3 and 54 cm3 respectively, AX : XB =
A. 2 : 1
B. 2 : 3
C. 8 : 19
D. 8 :27
E : 3 38
Answer
Answer is A
Help
To Q3

26. V D C A B M Q3
In the figure, VABCD is a right pyramid with a rectangular base. If AB=18cm, BC=24cm and CV=25cm, find
a) the height (VM) of
the pyramid,
b) volume of the
pyramid.
Answer
a) 20cm
b) 2880cm3
Help
To Q4

27. Q4 A C B
50cm
48cm
The figures shows a right circular cone ABC. If AD= 48cm and AC= 50cm, find
(a) the base radius (r) of the cone,
(b) the volume of the cone.
(Take  = ) 22 7
a) 14cm
b) 704cm3
Answer
Help

28. Answer for Q1 what is the length of a side of the base?
Let V is the volume of the pyramid and y be the length of a side of base
V =  base area  height 1 3
what is the length of a side of the base?
96 =  y2  8 1 3
8cm
288 = 8y2
Back to Q1
36 = y2
y = 6
To Q2
Therefore, the length of a side of base is 6 cm

29. Answer for Q2 AX : XB = ? Hints: Using the concept of RATIOS ( )3 = AB
( )3 = AB AX 16 54 A X Y B C AB AX
( )3 = 8 27 AX AB = 2 3
AB = AX + XB and AX = 2, AB = 3
Back to Q2
3 = 2 + XB
XB = 1
Therefore, AX : XB = 2 : 1
To Q3

30. Answer for Q3 AC2 =182 + 242 Volume of the pyramid is: AC2 = 900 1
a) the height (VM) of the pyramid
b) volume of the pyramid.
AC2 =
Volume of the pyramid is:
AC2 = 900
×base area ×height 3 1
AC = 30cm
MC = AC =15cm 2 1
= ×18 ×24 ×20 1 3
Answer for Q3
252 = VM2 + MC2
= 2880cm3
625 = VM
Therefore, the volume of the pyramid is 2880cm3
= VM2
VM2 = 400
Back to Q3
VM = 20cm
Therefore, the height (VM) of the pyramid is 20 cm
To Q4

31. Answer for Q4 (a) the base radius (r) (b) the volume of the cone
The volume (V) of cone is:
The radius is r, therefore:
V =  r2 h 3 1
502 = r2
=   142  48 3 1 22 7
2500 = r2
196 = r2
= 704 cm3
r = 14
The volume is 704 cm3 A
The radius is 14cm. C B
50cm
48cm
(Take  = 22/7)
Back to Q4

32. End of lesson!